Applications of the Mole
Molar Mass,
Percent Composition
&
Empirical and Molecular Formulas
Molar Mass
Molar mass is the term for the mass of
one mole.
 It may also be referred to as gram
molecular mass, gram formula mass,
and gram atomic mass.
 The unit is g/mol.

This photograph
shows one mole of
solid (NaCl)
liquid (H2O)
gas (N2).
Molar Mass and Elements
To determine the molar mass of an
element, find the element’s symbol on the
periodic table and round the mass to one
decimal.
 The molar mass of carbon is 12.0 g/mol, of
chlorine (Cl) is 35.5 g/mol and of iron (Fe)
is 55.8 g/mol.

Question

Calculate the molar mass of the each
of the following elements.
a) sulfur (S)
(32.1 g/mol)
b) chromium (Cr)
(52.0 g/mol)
c) bromine (Br)
(79.9 g/mol)
Molar Mass and
Compounds
To determine the molar mass of a
compound, find the mass of all
elements in the compound.
 If necessary, multiply an element’s
mass by the subscript appearing beside
that element in the compound’s formula
(or product of the subscripts).

Molar Mass and
Compounds Example

To determine the mass water (H2O), find
the mass of hydrogen and oxygen using
the periodic table.
Hydrogen: 1.0
Oxygen: 16.0
Molar Mass and
Compounds Example

Because there are 2 hydrogens in water
(H2O), multiply the mass of hydrogen by
two.
Hydrogen: 1.0 x 2 = 2.0
Oxygen: 16.0
Molar Mass and
Compounds Example

Add all numbers.
Hydrogen: 1.0 x 2 = 2.0
Oxygen: 16.0
18.0 g/mol
Question

Calculate the molar mass of each of the
following compounds.
a) Na2S
b) N2O4
c) C6H12O6
d) Ca(NO3)2
(78.1 g/mol)
(92.0 g/mol)
(180.0 g/mol)
(164.1 g/mol)
Interconverting Mass and Moles
 Mass can be converted to moles and moles
to mass by using dimensional analysis
(the factor label method).
Example grams to moles
Conversion

How many moles of lithium are in 2.00 g
of Li?
2.00 g Li
__
1
mol Li
mol Li
__
6.94 g Li
(0.288 mol Li)
Question

How many moles of sodium oxide are in
42.0 g of Na2O?
42.0 g Na2O
__
1
mol Na2O
__
62.0 g Na2O
(0.677 mol)
Question

How much would 3.45 moles of uranium
(U) weigh?
3.45 mol U
__
238.0
gU
__
1 mol U
(821 g)
Example Mass-Mole
Conversion

How many moles of magnesium are in
56.3 g of Mg?
56.3 g Mg
__
1 mol Mg
__
24.3
g Mg
(2.32 mol)
Question

How many moles is 5.69 g of NaOH?
5.69 g NaOH
__
1 mole NaOH
__
40.0
g NaOH
(0.142 mol)
Question

How many grams of sodium chloride
are in 3.45 moles of NaCl?
3.45 mol NaCl
__
58.5
g NaCl
__
1 mol NaCl
(202 g)
Questions

How many moles is 4.8 g of Cr(ClO3)2?
(0.0192 mol)

How many grams is 9.87 moles of H2O?
(178 g)
Questions

How many moles are in 6.8 g of CH4?
(0.425 mol)

How much does 49.0 moles of C6H12O6
weigh?
0.272 g
Percent Composition
Every compound has a definite
composition.
 This composition is usually stated as a
percent by mass of each element in the
compound.

Percent Composition

The % of an element, X, in a compound
can be found using the following formula:
(molar mass X)(#Xs)
%X =
x 100%
molar mass compound
Percent Composition
Determine the percent composition of
chlorine in calcium chloride (CaCl2).
1.The first equality statement comes
from the formula:
1 mol CaCl2 = 2 mol Cl2. Use the PToE to find the molar masses
of the constituent ions:
1molCa(40.1 g/mol) + 2molCl(35.5 g/mol) = 111. g
Percent Composition
Determine the percent composition of
chlorine in calcium chloride (CaCl2).
3. Use the data in the % composition
formula:
(molar mass Cl)(2Cl)
%Cl =
molar mass CaCl2
%Cl =
(35.5 g/mol)(2 molCl)
(111. g/mol)
= 63.9%
Percent Composition
1. What is the percent composition of
carbon in sodium acetate?
29.3%
Percent Composition
2. What is the percent composition of
aluminum in AlO3?
Percent Composition
3. What is the percent composition of
oxygen in Mg(NO3)2?
Percent Composition
4. What is the percent composition of sulfur
in Al2(SO4)3?
Empirical Formula
The empirical formula is the lowest wholenumber ratio of atoms of elements in a
compound.
 In many cases, the empirical formula is
the same as the true formula of the
compound.
 Ex. Sodium chloride, NaCl

Empirical Formula
Ex. glucose C6H12O6
 The subscripts in glucose are all divisible
by 6.
 The empirical formula is

– C(6/6)H(12/6)O(6/6)
– CH2O
Empirical Formula
Determine the empirical formula for:
1. Tl2C4H4O6
2.
N 2O 4
Empirical Formula
Percent composition data can be used to
determine the empirical formula of an
unknown compound.
 Example: The percent composition of an
unknown compound is found to be
38.43% Mn, 16.80% C, and 44.77% O.
Determine the empirical formula

Empirical Formula
Because percent means “per hundred,”
assume that you have 100g of a
compound.
 Based on your assumption, calculate the
number of moles of each element in 100 g
of the compound.

Empirical Formula
Find the lowest whole-number ratio by
dividing the moles of each element by the
smallest number.
 These numbers become the subscripts in
the empirical formula.
 Note: You may have to round, but these
numbers must be whole numbers.

Empirical formula
1. The composition of an unkown acid is 40.00% C, 6.71%
H, and 53.29% O. What is the empirical formula?
CH2O
Empirical formula
2. The composition of an unkown ionic compound is
60.7% Ni, and 39.3% F. What is the empirical formula?
NiF2
Molecular Formula
For ionic compounds, the empirical
formula and the formula unit are the
same.
 For molecules (covalent bonding), the
empirical formula may not be the true
formula.

Molecular Formula
The molecular formula tells the exact
number of atoms of each element in a
molecule.
 The molecular formula can be the same
as the empirical formula, or it will be a
whole-number multiple of the empirical
formula.

Molecular Formula



Unlike the empirical formula, the molecular
formula of an unknown compound cannot be
determined from the % composition alone.
In addition to the empirical formula, one must
know the molar mass of the compound.
Then the molar mass of the compound can
be compared with the molar mass of the
empirical formula.
Molecular Formula
1. The molar mass of benzene is 78 g/mol, and its
emprical formula is CH. What is the molecular formula
for benzene?
1st, calculate the molar mass of the emprical formula, CH.
Then, divide the molar mass of benzene by the molar
mass of the empircal formula to get the multiple, n.
Finally, multiply the whole-number multiple, n, by the
subscripts in the emprical formula to get the molecular
formula:
Molecular Formula
2. The empirical formula for butane is C2H5, and its molar
mass is 60.0 g/mol. What is the molecular formula for
butane?
Rounding
When determining the empirical and molecular formulas, a
certain amount of rounding is necessary.
Round to these values, and multiply by the LCD to get the
lowest whole-number ratio:
1: no multiplication necessary
0.5 (1/2), multiply by 2
0.33/0.66 (1/3 & 2/3), multiply by 3
0.25/0.75 (1/4 & ¾), multiply by 4
0.2/0.4/0.6/0.8 (1/5, 2/5, 3/5 & 4/5), multiply by 5.